Download NEGATIVE FEEDBACK and APPLICATIONS

PH-315
Portland State University
A. La Rosa
NEGATIVE FEEDBACK and APPLICATIONS
I. PURPOSE:
II. THEORETICAL CONSIDERATIONS
II.1 Negative feedback
II.2 Operational Amplifiers with “Infinite” and Finite Gain
II.2.A The concept of open loop gain and close loop gain
II.2.B Op-amps with infinite open loop gain:
The Golden Rules of Operational Amplifiers
III. EXPERIMENTAL CONSIDERATIONS
Applications of negative feedback with operational amplifiers
III.1 Amplifiers Circuits
f3dB frequency (test in the frequency domain)
Slew-rate (test in the time domain)
Transfer function (analysis in the Laplace domain)
III.2 Active Filters
III.2.A Exploiting the low output impedance and high input impedance of op-amps
III.2.B Active pass filters: Filters with steeper roll-off
I. PURPOSE:
To use various types of negative feedback, using operational amplifiers, to build a) gaincontrolled amplifier, non-inverted amplifier, integrator, differentiator; and b) active
filters.
Students must i) provide an analysis of the circuit in the time domain, and ii) evaluate
the corresponding transfer function (Laplace domain analysis) of the circuits.
II. THEORETICAL CONSIDERATIONS
FEEDBACK
In control systems, feedback consists in comparing the output of the system with the desired
output and making a correction accordingly.1
II.1 Negative feedback
Negative feedback is the process of coupling a portion of the output back into the input, as
a way to cancel part of the input.
This process, it turns out, has the effect of reducing the gain of the amplifier, but, in
exchange, it improves other characteristics including freedom from distortion and
nonlinearity, flatness in the frequency response, and predictability. In fact, as more negative
feedback is used, the resultant amplifier’s characteristics becomes less dependent on the
characteristics of the original open-loop amplifier
Caution: We have learned from our RC filter experiment that a phase lag exists between the
input and output voltages, which increases as more components are added into the circuit.
If a negative feedback-loop circuit were to accumulate large enough phase lag (i.e. greater
than 180o), then positive feedback occurs (the circuit ends up being an oscillator.)
vout
Differential
input voltage
vv+
-
+VCC
ii+
vout =
~ 120 V
-VCC /A
= A (v+ - v- )
+VCC /A
+
(v+ - v- )
-VCC
Ri > 10 M
i- , i+
~ 20 nA for the LM358AP
~ pA for op amps with FET-input types
Amplifying
region
Fig. 1 Operational amplifier.
II.2 OPERATIONAL AMPLIFIERS WITH “INFINITE and FINITE GAIN

See additional information contained in the file “Operational Amplifiers (background)”
posted on the website of this course.

Before performing the experiments listed below, review the data sheet of the op amp you
are using. In particular, check the slwe rate (it is 0.3 V/s for the LM358AP). This parameter
(that tell you how fast tye op-amp responds to an input signal) will allow you to estimate
the frequency bandwidth response of your circuit.
Voltage Gain (in dB)
II.2.A The concept of open loop gain and close loop gain
Gain (in dB)
100
Open loop
voltage gain
80
60
vin
+
A ()
40
f
20
0
vout
A
vout 
A=
3dB
0
10
Low-pass_unity-gain-Bandwidth
-
10
1
10
2
10
3
10
4
10
5
10
Frequency
6
10
vin 
Open loop
voltage gain
7
Unity-gain
ffTT Unity-gain
bandwidth
bandwidth
Fig. 2 Typical open loop voltage gain A as a function of frequency.
When integrated with an external network, the gain of the network is less than the open-loop
gain of the amplifier. (Even though the gain is lower, the performance in terms of stability is
much better).
Voltage Gain (in dB)
Open loop
voltage gain
60
40
20
ACL
Ideal close
100
80
Gain (in dB)
loop loop
gain:
A () Open
3
Low-pass_unity-gain-Bandwidth
Rf = 103 k
ACL=10
voltage gain
ACL ()
Ro =
1 k
vin
Close loop
f
voltage3dB
gain
0
10
0
10
1
10
2
10
3
10
Frequency
Close loop bandwidth
4
10
5
10
6
10
7
A
vout
vout 
ACL = v 
in
fT Unity-gain
Fig. 3 Open loop A and close loop ACL voltage
gain as a function of frequency. Notice in the
bandwidth
graph that ACL < A.
II.2.B Op-amps with infinite open loop gain:
The Golden Rules of Operational Amplifiers
When implemented as part of a negative feedback external network, the behavior of the op
amp can be predicted (for many practical purposes) based on two simple rules: 2
Rule I
The output attempts to do whatever is necessary as to
produce that the external feedback brings the differential
input voltage close to zero (v+ - v- ≈ 0).
(1)
The inputs draw no current (i- , i+ ≈ 0).
Rule II
Basically we are saying that the open loop gains A is infinite. This is of course an abstraction, but
allows obtaining a quick grasp of the functioning of many widely used circuits (as we will see in
the experimental section below).
III. EXPERIMENTAL CONSIDERATIONS
Applications of negative feedback with operational amplifiers
III.A Amplifiers Circuits
III.B Active Filters
III.1 Amplifiers Circuits
III.1.A The Inverting Amplifier
Since v+ is grounded, then Rule I implies v- = 0. Accordingly,
the current through Ro is equal to Io =Vin / Ro; and
the current through Rf is then equal to If = -Vout / Rf.
Since no current flows into the op amp inputs (Rule II) we should have
Io = If
that is,
Vin / Ro= -Vout / Rf,
or simply,
Vout  
Rf
Ro
Vin
(2)
TASKS:
a) Construct the circuit shown in Fig. 4, and verify that indeed the voltage gain is equal to
R f / Ro .
b) Testing the operational amplifier in the frequency domain. Using a sinusoidal wave (~
50 mV amplitude) as an input, make two Bode plots, corresponding to two different
gains (10 and 100 for example).
Evaluate the corresponding f3dB for these two cases. (i.e. evaluate if the f3dB value
depends on the gain.)
c) Testing the operational amplifier in the frequency domain. This step is to get familiar
with the concept of slew-rate [see also the “slew-rate” section in the file “Operational
amplifier (background material)” posted online in the website of this course]
For a given value of the slew-rate S, we expect the maximum amplitude at which a sinewave output of frequency f remains undistorted to be given by S= 2f Bmax.
That is, the amplitude B of a sine-wave B Sin(t) must fulfill the condition
S
B
(3)
2 f
c1) Using first a square-wave input of 50 mV amplitude, increase gradually the frequency
until you start noticing that the output signal does not follow exactly the input. You
are this way testing the limitation of the operational amplifier in the time domain
(rather than in the frequency domain, as in part b above).
c2) Change you input to a sinusoidal wave, and use a fixed frequency f that is lower
than, but close to, the 3dB found in part b above. Check the amplitude of the output
signal has the expected value (for the given gain, 10 or 100, you are using).
Now take different (increasing) values of the input voltage amplitude B. Verify if the
amplitude of the output-signal deviates very much from its expected value when the
input-signal amplitude grows too large. Check if this deviation start happening when
B starts exceeding the value Bmax=S/ 2 f (where S is the slew rate you obtain from
the data sheet of the op-amp). Make this evaluation for the two cases, gain 10 and
100.

Rf = 10 k
If
Vin
Ro = 1 k
Io
vv+
+
Vout
VCC = 12V
-VCC = -12V
Fig. 4 DC inverted amplifier circuit.
d) We analyzed above the functioning of the inverted amplifier circuit in the time domain.
This time, instead, we ask you to also analyze the circuit in the Laplace domain; evaluate
the transfer function of this circuit.
Suggestion: Proceed with a time-domain analysis first, and then evaluate the
corresponding Laplace transformation of the quantities involved.
Rf = 10 k
if
vin
Ro = 1 k
vv+
io
vout
+
VCC = 12V
-VCC = -12V
Fig. 5 AC inverted amplifier circuit. (It is the same circuit as in Fig. 4;
it just emphasizes the ac input and output voltages.
III.1.B The Noninverting Amplifier
Rule I implies v- = Vin
At the same time, v- is part of a voltage divider: v- = [Vout /( R1+ R2)] R1
Equating these two expressions, we obtain,
 R 
Vout  1  2  Vin
 R1 
Vin
+
(4)
Vout
R2 = 20 k
R1 = 2 k
Fig. 6 Non-inverting amplifier.
Implementation:
a) Construct the circuit shown in Fig, 6, whose input is a DC voltage, and verify that the voltage
gain is indeed equal to 1 + (R2 / R1). Using a square wave (~50 mV amplitude) as an input,
make two Bode plots, corresponding to two different gains (10 and 100 for example).
Check if the f3dB value depends on the gain.
b) Analyze the circuit in the Laplace domain; evaluate the transfer function of this circuit.
Suggestion: Proceed with a time-domain analysis first, and then evaluate the corresponding
Laplace transformation of the quantities involved.
III.1.C Differential Amplifier
(SKIP THIS SECTION IN 2004)
Implementation:
a) Apply the golden rules to demonstrate that the output voltage of the circuit shown in Fig. 5 is
given by,
Vout 
R2
( V2  V1 )
R1
(5)
b) Build the circuit shown in Fig, 7 and verify if the output voltage varies according to the
expression given in part a) above.
R2
R1
V1
-
R2 = 20 k
R1 = 2 k
Vout
+
R1
V2
R2
Fig. 7 Differential amplifier.
III.1.D Integrator
Op amps allow to make integrators without the restriction that Vout < Vout .(as required
when using only R and C components.)
Since v + is grounded, the input v- acts as a virtual ground. The current IR passing through the
resistor is then given by,
IR = Vin /R
The current through the capacitor is given by,
d
d
I C  (q)  (0  Vout )C
dt
dt
 C
dVout
dt
Rule II implies that IR = IC ,
Vin
dV
 C out
R
dt
This implies,
Vout (t )  
1 t
Vint (t ' )dt '
RC 
(6)
C= 0.01 F
q -q
IC
R = 1 k
vin
IR
+
vout
VCC = 12V
-VCC = -12V
Fig. 8 Integrator.
Implementation:
a) Implement an integrator circuit shown in Fig. 8. Since charging effects can cause serious
offsets, a parallel resistor Rp may be needed (to prevent any long term voltage shift at the
input). Try different values for Rp. (100K, 1 M). See Fig. 9.
b) Test your circuit using a square signal of 1 kHz at the input. Investigate the effect of changing
the various parameters.
Rp
C= 0.01 F
Vin
R = 1 k
Vout
+
VCC = 12V
-VCC = -12V
Fig. 9 Integrator.
c) Analyze the circuit in the Laplace domain; evaluate the transfer function of this circuit.
Suggestion: Proceed with a time-domain analysis first, and then evaluate the corresponding
Laplace transformation of the quantities involved.
III.1.E Differentiator
The circuitry is similar to the integrator but with the R and C reversed.
The current through the capacitor is given by,
d
d
(q)  (VintC )
dt
dt
dV
 C in
dt
IC 
The current through the resistor is given by,
IR = (0-Vout /R
Rule II implies that IR = IC ,
C
dVin
V
  out
dt
R
This implies,
Vout (t )   RC
dVin
dt
(7)
Implementation:
a) Implement a differentiator circuit. Test the circuit with triangle waves at the input.
R = 1 k
Vin
C= 0.01 F
q -q
IR
-
Vout
+
IC
VCC = 12V
-VCC = -12V
Fig. 10 Differentiator
b) We analyzed above the functioning of the circuit in the time domain. This time, instead,
analyze the circuit in the Laplace domain; evaluate the transfer function of this circuit.
III.2 Active Filters
III.2.A Exploiting the low output impedance and high input impedance of op-amps
In a previous laboratory you worked with two first-order low pass filters connected in a cascade
arrangement, as shown in Fig. 11.
vin
R
R
vout
C
C
Fig. 11 Two low-pass RC filters.
The output voltage of this circuit is given by,
zC
zC
1
vout  vin
( R  z C ) ( R  z C ) 1  R z C / ( R  z C )2
where z C is the capacitor impedance.
(8)
Notice, we could not have stated that vout was simply the product of two voltage
zC
zC
dividers expressions like
because of the “loading” effects of
(R  zC ) (R  zC )
the second simple low-pass filter over the first simple low-pass filter.
TASK: Using expression (8) prove that,
1
vout 
vin
( jRC  1)2  jRC
(9)
One way to get around the loading effect one circuit stage exerts on another is to use
operational amplifiers in a “follower” configuration. Notice, it is simply a non-inverting
amplifier, as shown in Fig. 6 above with R2=0 and R2=∞).
TASKS: Build the circuit shown in Fig. 12.
Make a Bode plot and identify the corresponding f3dB frequency. Compare the result
with the one corresponding to the circuit in Fig. 11 (obtained in your previous lab-4).
Evaluate vout using an analysis in the time domain (use the op-amp golden rules).
Evaluate the corresponding transfer function (analysis in the Laplace domain).
In both cases justify all your steps.
vin
R
+
C
R
-
vout
C
IR
Fig. 12 Low pass filters with a decoupling amplifier.
Hint:
The answers you should arrive are the following,
vout
zC
zC

(R  zC ) (R  zC )
vin
(10)
In the time domain z C  1 / ( j C ) ; hence,
1
1
vout (t )

(1  jRC ) (1  jRC )
vin (t )
(11)
In the time domain z C  1 / (C s) ; hence,
1
1
Vout (s)

(1  RC s) (1  RC s)
Vin (s)
(12)
III.2.B Active pass filters: Filters with steeper roll-off
Simple RC filters produce gentle low low-pass gain characteristics with a roll-off of
6dB/octave at frequencies well beyond the -3dB frequency. Such filters are sufficient for many
multiple purposes. Often, however, filters with flatter pass-bands and steeper roll-off are
needed.
One solution may consists of cascading many RC filters, using buffer amplifiers (as indicated in
Fig. 12) to avoid loading effects. It turns out, such an approach produces indeed steeper roll-off,
but the curved “f3dB knee” in the gain vs frequency response does not disappear.
On the other hand, filters made with inductors and capacitors can have very sharp
responses. In fact, by including inductors in the design, it is possible to create filters with any
desired flatness of passband, combined with roll-off steepness. The only problem is that
inductors as circuit elements are often bulky and expensive (not to mention having significant
series resistance. Hence the task becomes to design “inductorless filters” with the
characteristics of ideal RLC filters.
Solution: Using op-amps as part of the filter design, it is possible to synthesize any RLC filter
characteristic without using inductors.
A specific example of this statement is provided in the next example, where you are
asked to compare the gain, as well as the transfer functions, corresponding to a RLC
passive filter and a active RC filter.
TASKS: Consider the two circuits shown in Fig. 13.
a) For the passive RLC filter:
a1) Using the method of complex impedance in the time domain, show that
vout (t )
1

2
vin (t )
-  LC  jRC  1
(13)
a2) Using the method of complex impedance in the Laplace domain, show that,
Vout (s)
1

2
Vin (s)
LC s  RC s  1
(14)
Hint: Use the expressions, in the Laplace domain, for the individual impedances
of the R, C and L elements, as given in the “Transfer Function” notes posted in
the website of this course.
vin
L
vout
R
C
vin
R
R
C
Fig. 13 Left: RLC passive filters. Right: Active RC pass filter.
vout
+
C
IR
b) For the active RC filter shown in Fig. 13:
b1) Show that
vout
zC
zC

(R  zC ) (R  zC )
vin
(15)
b2) Show that in the time domain, where z C  1 / ( j C ) , the expression above
becomes,
1
1
vout (t )
(16)

(1  jRC ) (1  jRC )
vin (t )
Show that in the Laplace domain, where z C  1 / (C s) , the expression above
becomes,
Vout (s)

Vin (s)
1
1
1

2 2
(1  RC s) (1  RC s)
( RC ) s  2 RC s  1)
(17)
Hint: Use the expressions, in the Laplace domain, for the individual impedances
of the R, C and L elements, as given in the “Transfer Function” notes posted in
the website of this course.
c) Compare the transfer functions for the passive RLC circuit (given by expression (14))
and the transfer function for the active RC filter (given by expression (17)). Include
your comments on whether this active RC pass filter will behave similar to a passive
RLC circuit.
1
2
(Chapter 4) Horowitz and Hills, “The Art of Electronics.” 2nd Ed.; Cambridge University Press (1990).
(Page 177) Horowitz and Hills,